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Chapter 26: Problem 73

The inner conductor of a coaxial cable has a radius of \(0.800 \mathrm{mm},\) and the outer conductor's inside radius is \(3.00 \mathrm{mm} .\) The space between the conductors is filled with polyethylene, which has a diclectric constant of 2.30 and a dielectric strength of \(18.0 \times 10^{6} \mathrm{V} / \mathrm{m} .\) What is the maximum potential difference that this cable can withstand?

Short Answer

Expert verified
The maximum potential difference that this coaxial cable can withstand is 39600 Volts.

Step by step solution

01

Understand and define given parameters

The following parameters are given in the problem: the inner radius of the cable (\(r_i = 0.800\) mm), the outer radius of the cable (\(r_o = 3.00\) mm), and the dielectric strength of the material between the conductors (\(E_{max} = 18.0 * 10^6\) V/m). The radii are given in mm, so those should be converted to meters before beginning with the calculations, giving us \(r_i = 0.0008\) m, \(r_o = 0.003\) m.
02

Calculate the maximum potential difference

The maximum potential difference (\(V_{max}\)) can be calculated using the formula for electric potential difference in a cylindrical capacitor with the dielectric, which is \(V_{max} = E_{max} * d\). Here, \(E_{max}\) is the dielectric strength and \(d\) is the distance between the conductors. The distance \(d\) is equal to \(r_o - r_i\) which is \(0.003 m - 0.0008 m = 0.0022 m\). So, \(V_{max} = 18.0 * 10^6 V/m * 0.0022 m = 39600 V\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Coaxial Cable
A coaxial cable is a type of electrical cable consisting of two concentric conductors. The central conductor is surrounded by a tubular insulating layer, typically made of a dielectric material. This is further enclosed by a conductive shield and, finally, an insulating outer jacket. Such cables are widely used for transmitting electrical signals and are commonplace in networking and television setups.

Understanding the structure of coaxial cables is crucial for grasping their functionality:
  • Inner Conductor: This carries the electrical signal forward. It is the main path of current flow.
  • Outer Conductor (Shield): This acts as a return path for the signal and also provides shielding against electromagnetic interference.
  • Dielectric Insulation: The space between the inner and outer conductors is filled with this insulating material.
One key feature of coaxial cables is their ability to deliver stable and secure signal transmission over distances, thanks to their design.
Dielectric Materials
Dielectric materials are insulating substances that greatly affect the behavior of electric fields within them. When introduced into a capacitor or used in wiring like in coaxial cables, these materials can withstand electric fields without conducting electricity.

One significant property of dielectric materials is the dielectric constant. This dimensionless measure determines how effectively the material can store electrical energy compared to vacuum. For example, polyethylene's constant is 2.30, enhancing the cable's capacitance compared to air.
  • Dielectric Strength: This is the maximum electric field a material can endure without breakdown. For polyethylene, it鈥檚 18.0 x 106 V/m, meaning it can handle substantial electric fields before failing, making it suitable for high-voltage applications.
Understanding dielectric properties is essential for designing circuits and choosing materials that can handle specific electrical stresses.
Electric Potential Difference
Electric potential difference, commonly referred to as voltage, represents the work needed to move a unit charge between two points in an electric field. In the context of a coaxial cable, this refers to the difference in electric potential between the inner and outer conductors.

In our example, the formula for calculating this difference in a cylindrical setup involves multiplying the dielectric strength by the distance between the conductors:
  • The extbf{distance} is the radial difference, found by subtracting the inner radius from the outer ( _o - _i ight).
  • The extbf{maximum potential difference}, or the highest potential the cable can handle before dielectric breakdown occurs, is given by ext{V}_{max} = E_{max} imes d.
Understanding potential difference is vital as it ensures that the cable operates safely without reaching the breakdown point.

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Most popular questions from this chapter

According to its design specification, the timer circuit delaying the closing of an elevator door is to have a capacitance of \(32.0 \mu \mathrm{F}\) between two points \(A\) and \(B .\) (a) When one circuit is being constructed, the inexpensive but durable capacitor installed between these two points is found to have capacitance \(34.8 \mu \mathrm{F}\). To meet the specification, one additional capacitor can be placed between the two points. Should it be in series or in parallel with the \(34.8-\mu \mathrm{F}\) capacitor? What should be its capacitance? (b) What If? The next circuit comes down the assembly line with capacitance \(29.8 \mu \mathrm{F}\) between \(A\) and \(B .\) What additional capacitor should be installed in series or in parallel in that circuit, to meet the specification?

Two conducting spheres with diameters of \(0.400 \mathrm{m}\) and \(1.00 \mathrm{m}\) are separated by a distance that is large compared with the diameters. The spheres are connected by a thin wire and are charged to \(7.00 \mu \mathrm{C}\). (a) How is this total charge shared between the spheres? (Ignore any charge on the wire.) (b) What is the potential of the system of spheres when the reference potential is taken to be \(V=0\) at \(r=\infty ?\)

Capacitors \(C_{1}=6.00 \mu \mathrm{F}\) and \(C_{2}=2.00 \mu \mathrm{F}\) are charged as a parallel combination across a \(250-\mathrm{V}\) battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate. Calculate the resulting charge on each capacitor.

A small rigid object carries positive and negative \(3.50-\mathrm{nC}\) charges. It is oriented so that the positive charge has coordinates \((-1.20 \mathrm{mm}, 1.10 \mathrm{mm})\) and the negative charge is at the point \((1.40 \mathrm{mm},-1.30 \mathrm{mm}) .\) (a) Find the electric dipole moment of the object. The object is placed in an electric field \(\quad \mathbf{E}=(7800 \mathbf{i}-4900 \hat{\mathbf{j}}) \mathrm{N} / \mathrm{C} .\) (b) Find the torque acting on the object. (c) Find the potential energy of the object-ficld system when the object is in this orientation. (d) If the oricntation of the object can change, find the difference between the maximum and minimum potential energies of the system.

(a) Two spheres have radii \(a\) and \(b\) and their centers are a distance \(d\) apart. Show that the capacitance of this system is $$C=\frac{4 \pi \epsilon_{0}}{\frac{1}{a}+\frac{1}{b}-\frac{2}{d}}$$ provided that \(d\) is large compared with \(a\) and \(b\). (Suggestion: Because the spheres are far apart, assume that the potential of each equals the sum of the potentials due to each sphere, and when calculating those potentials assume that \(V=k_{e} Q / r\) applies.) \((\mathrm{b})\) Show that as \(d\) approaches infinity the above result reduces to that of two spherical capacitors in series.
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